Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$

Question

We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an $N' \in \mathbb{N}$ such that $\left| \frac{1}{x_n} - \frac{1}{2} \right| < \epsilon'$ for all $n \geq N'$.

Let $\varepsilon = \varepsilon' - \frac{3}{2}$. Then by the triangle inequality, we have $$\left| \frac{1}{x_n} - \frac{1}{2} \right| \leq \left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right|.$$

Because we proved earlier that $x_n > 1$ for all $n \geq N$, we know that $$\left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right| < \left| x_n\right| + \left| -\frac{1}{2}\right|.$$

By the triangle inequality again, we know that $$\left| x_n - \frac{1}{2}\right| \leq \left| x_n\right| + \left| -\frac{1}{2}\right|.$$

Note that $\left| x_n - \frac{1}{2}\right| = \left| (x_n - 2) + (2 - \frac{1}{2})\right|$, so we get $$\left| (x_n - 2) + (2 - \frac{1}{2})\right| \leq \left| x_n - 2\right| + \left| 2 - \frac{1}{2}\right|.$$

Then we have $$\left| x_n - 2\right| + \left|2 - \frac{1}{2}\right| < \varepsilon + \frac{3}{2} = \left(\varepsilon' - \frac{3}{2}\right) + \frac{3}{2} = \varepsilon',$$ so $\frac{1}{x_n} \rightarrow \frac{1}{2}$.

I just realized that I don't know how to make sure that $\varepsilon' - \frac{3}{2}$ is a positive number. How can I do this?

Answer 1

Note that $$\left|\frac{1}{x_n}-\frac{1}{2}\right|= \left|\frac{2-x_n}{2x_n} \right|$$ Choose $N$ large enough that $x_n>1$ and $|x_n-2|<\epsilon$ for all $n>N$. Then $$ \left|\frac{2-x_n}{2x_n}\right|\leq \epsilon/2< \epsilon $$ for all $n>N$.

Answer 2

In fact there's no reason why $\varepsilon' - \frac{3}{2}$ should be positive.

Instead notice that for all natural numbers $n$ such that $x_n\neq 0$, it holds that $$\left|\dfrac{1}{x_n}-\dfrac1 2\right|=\left|\dfrac{x_n-2}{2x_n}\right|.$$

Relating the RHS to the convergence of $(x_n)_{n\in \mathbb N}$ and to $(x_n)_{n\in \mathbb N}$ being greater (eventually) than $1$ should get you there.

Answer 3

Use the Algebraic Limit Theorem for sequences:

1. $\lim(ca_n)=c\cdot\lim a_n$
2. $\lim(a_n+b_n)=\lim a_n + \lim b_n$
3. $\lim(a_n b_n)=\lim a_n\cdot\lim b_n$
4. $\lim\left(\frac{a_n}{b_n}\right)=\frac{\lim a_n}{\lim b_n}$ provided $b_n \ne 0$ and $\lim b_n \ne 0$

Rule (4) in the backwards direction, with $(a_n) = (1, 1, 1, \dots)$ and $(b_n) \to 2$, will give you what you need. If $\lim a_n = 1$ and $\lim b_n = 2$, then $\lim\left(\frac{1}{b_n}\right) = \frac{1}{2}$.

A proof of the Algebraic Limit Theorem can be found on p. 45 of this book:

Abbott, S. (2001). Understanding analysis. New York, NY: Springer.